Mark the correct alternative in each of the following:

Let g(x) = 1 + x – [x] and where [x] denotes the greatest integer less than or equal to x. Then for all x, f (g(x)) is equal to


Given that g(x) = 1 + x – [x] and



where [x] denotes the greatest integer less than or equal to x.


(i) -1 < x <0


g(x) = 1 + x – [x]


g(x) = 1 + x + 1 { [x] = -1}


g(x) = 2 + x


f(g(x))= f(2 + x)


f(g(x))=1 + 2 + x-[2 + x]


f(g(x)) = 3 + x -2 – x


f(g(x)) = 1


(ii) x = 0


f(g(x)) = f(1 + x-[x])


f(g(x)) = 1 + 1 + x –[x] – [1 + x + [x]]


f(g(x)) = 2 + 0 -1


f(g(x)) = 1


(iii) x > 1


f(g(x)) = f(1 + x-[x])


f(g(x)) = f(x>0) = 1


Hence, f(g(x)) = 1 for all cases.

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